當int型別超出了[-2147483648,2147483647]會發生什麼?
阿新 • • 發佈:2018-12-08
當int超超出範圍,編譯器會報錯吧?。。。。
答案是不會(看編譯器怎麼處理),在vs環境中,當輸入下面的程式碼:
int a = 2147483647;
int b = 2147483648;
int d = 2147483649;
int e = 2147483650;
結果是:
也就是說2147483648變成了-2147483648,即迴圈賦值了,一旦超出就迴圈到最小的值,依此類推。
那麼在計算中怎麼判斷溢位呢,比如,有個迴圈不停累加,當溢位就返回一個結果。
我們用一個更大的數,比如__int64(兩個32位數再怎麼加、乘運算也不會溢位64位,比如4位的1111b乘以1111b(最高位為符號位)就是十進位制的7*7,最大的也8位在-128~127之間 )
有人說用long,long這個慎用,因為在32位機器和64位機器上表示範圍不一樣,具體可以通過下面的方法查詢:
printf("int取值範圍:%d ~ %d\n", INT_MIN, INT_MAX);
printf("long取值範圍:%d ~ %d\n", LONG_MIN, LONG_MAX);
標頭檔案是:
#include <limits.h>
或者直接開啟這個標頭檔案:(我這是32位的機器)
// // limits.h // // Copyright (c) Microsoft Corporation. All rights reserved. // // The C Standard Library <limits.h> header. // #pragma once #define _INC_LIMITS #include <vcruntime.h> _CRT_BEGIN_C_HEADER #define CHAR_BIT 8 // number of bits in a char #define SCHAR_MIN (-128) // minimum signed char value #define SCHAR_MAX 127 // maximum signed char value #define UCHAR_MAX 0xff // maximum unsigned char value #ifndef _CHAR_UNSIGNED #define CHAR_MIN SCHAR_MIN // mimimum char value #define CHAR_MAX SCHAR_MAX // maximum char value #else #define CHAR_MIN 0 #define CHAR_MAX UCHAR_MAX #endif #define MB_LEN_MAX 5 // max. # bytes in multibyte char #define SHRT_MIN (-32768) // minimum (signed) short value #define SHRT_MAX 32767 // maximum (signed) short value #define USHRT_MAX 0xffff // maximum unsigned short value #define INT_MIN (-2147483647 - 1) // minimum (signed) int value #define INT_MAX 2147483647 // maximum (signed) int value #define UINT_MAX 0xffffffff // maximum unsigned int value #define LONG_MIN (-2147483647L - 1) // minimum (signed) long value #define LONG_MAX 2147483647L // maximum (signed) long value #define ULONG_MAX 0xffffffffUL // maximum unsigned long value #define LLONG_MAX 9223372036854775807i64 // maximum signed long long int value #define LLONG_MIN (-9223372036854775807i64 - 1) // minimum signed long long int value #define ULLONG_MAX 0xffffffffffffffffui64 // maximum unsigned long long int value #define _I8_MIN (-127i8 - 1) // minimum signed 8 bit value #define _I8_MAX 127i8 // maximum signed 8 bit value #define _UI8_MAX 0xffui8 // maximum unsigned 8 bit value #define _I16_MIN (-32767i16 - 1) // minimum signed 16 bit value #define _I16_MAX 32767i16 // maximum signed 16 bit value #define _UI16_MAX 0xffffui16 // maximum unsigned 16 bit value #define _I32_MIN (-2147483647i32 - 1) // minimum signed 32 bit value #define _I32_MAX 2147483647i32 // maximum signed 32 bit value #define _UI32_MAX 0xffffffffui32 // maximum unsigned 32 bit value // minimum signed 64 bit value #define _I64_MIN (-9223372036854775807i64 - 1) // maximum signed 64 bit value #define _I64_MAX 9223372036854775807i64 // maximum unsigned 64 bit value #define _UI64_MAX 0xffffffffffffffffui64 #if _INTEGRAL_MAX_BITS >= 128 // minimum signed 128 bit value #define _I128_MIN (-170141183460469231731687303715884105727i128 - 1) // maximum signed 128 bit value #define _I128_MAX 170141183460469231731687303715884105727i128 // maximum unsigned 128 bit value #define _UI128_MAX 0xffffffffffffffffffffffffffffffffui128 #endif #ifndef SIZE_MAX #ifdef _WIN64 #define SIZE_MAX _UI64_MAX #else #define SIZE_MAX UINT_MAX #endif #endif #if __STDC_WANT_SECURE_LIB__ #ifndef RSIZE_MAX #define RSIZE_MAX (SIZE_MAX >> 1) #endif #endif _CRT_END_C_HEADER
可以看出 long long和__int64是一樣的。
說到這,還有個問題:下面的a等於多少?
long long a = 2147483647 +1;
你可能覺得是2147483648,但結果卻是-2147483648,why?。。。
因為2147483647是在int範圍裡,編譯器把它當作int來處理,加1後結果還是int,超範圍了。
這樣結果就是:2147483648
long long a = (long long)2147483647 +(long long)1;